Answer:
P = 2 pi R / v period of space station
F / m = v^2 / R centripetal force per unit of mass
So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2
Also, F / m = 9.8 m/s^2 earth's gravitational attraction
So 9.8 = 4 pi^2 R / P^2 or R = 9.8 P^2 / 4 * pi^2) = 195 m
Or D = 2 R = 390 m the diameter required
<span>two objects in contact with each other are the same temperature</span>
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
A
More concentrated means more collisions per unit volume, and as the volume stays the same and only concentration changes, the there are more collisions